Answers
9514 1404 393
Answer:
(x, y) = (5.76, 1 5/7)
Explanation:
The location of the centroid in the x-direction is the ratio of the first moment of area about the y-axis to the total area. Similarly, the y-coordinate of the centroid is the first moment of area about the x-axis, divided by the area.
For the moment about the y-axis, we can define a differential of area as ...
dA = (y2 -y1)dx
where y2 = √(x/k2) and y1 = k1·x^3
The distance of that area from the y-axis is simply x.
So, the x-coordinate of the centroid is ...
[tex]\displaystyle c_x=\frac{a_x}{a}=\frac{\int{x\cdot dA}}{\int{dA}}\\\\a_x=\int_0^{12}{x(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{5k_2^{1/2}}\cdot12^{5/2}-\frac{k_1}{5}12^5\\\\a=\int_0^{12}{(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{3k_2^{1/2}}\cdot12^{3/2}-\frac{k_1}{4}12^4\\[/tex]
For k1 = 4/12^3 and k2=12/4^2, these evaluate to ...
[tex]a_x=115.2\\a=20\\c_x=5.76[/tex]
The y-coordinate of the centroid requires we find the distance of the differential of area from the x-axis. We can use (y2 +y1)/2 for that purpose. Then the y-coordinate is ...
[tex]\displaystyle c_y=\frac{a_y}{a}\\\\a_y=\int_0^{12}{(\frac{y_2+y_1}{2}(y_2-y_1))}\,dx=\frac{1}{2}\int_0^{12}{(\frac{x}{k_2}-(k_1x^3)^2)}\,dx\\\\a_y=\frac{12^2}{4k_2}-\frac{k_1^212^7}{14}=\frac{240}{7}\\\\c_y=\frac{12}{7}\approx1.7143[/tex]
The centroid of the shaded area is ...
(x, y) = (5.76, 1 5/7)
Related Questions
How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?
ASAP PLS
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Answer:
Explanation:
Work, U, is equal to the force times the distance:
U = F · r
Force needed to lift the weight, is equal to the weight: F = W = m · g
so:
U = m · g · r
= 20.4kg · 9.81 [tex]\frac{N}{kg}[/tex] · 1.50m
= 35.316 [tex]\frac{N}{m}[/tex]
= 35.316 W
a storage tank contains liquid with a density of 0.0361 lbs per cubic inch. the height of liquid in the tank is 168 feet. what is the pressure of the liquid at the bottom of the tank give your answer in psi
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Answer:
Objects that float on water have densities less than the density of water; those that do not float on water have densisties greater than the density of water:
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Do not float on water: d > 0.0361 lb/in³
Explanation:
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Answer:
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In the engineering design and prototyping process, what is the advantage of drawings and symbols over written descriptions?
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The advantages that can be associated to
drawings and symbols over written descriptions in engineering design and prototyping process are;
Communicate design ideas as well as technical information to engineers.
Symbols and drawings can be universal which means it is easy to interpret any where by professionals.
An engineering drawing serves as complex dimensional object and symbol use by engineer to communicate.Drawings and symbols makes it easier to communicate design ideas and technical information to engineers and and how the process will go.Therefore, drawings and symbols is universal to all engineer unlike written one.
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Answer:
hi
Explanation:
Should you recommend changing fluids based on the color or smell of the fluid?
Only if the color is brown and the fluid smells burned
Yes, always.
No. Stick with the manufacturer's recommendations.
Only if the fluid is red and smells sweet
Fire properties of the glass? |
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Answer:
Ordinary window glass cannot withstand the high temperatures associated with a structure fire, and it will break and fall out of its frame at about 250° F, only a few minutes into a fire. On the other hand, glass that is classified as "fire-rated" for at least 60 minutes can tolerate heat in excess of 1600° F
Determine the initial void ratio, the relative density and the unit weight (in pounds per cubic foot) of the specimens for each sand.
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The initial void ratio is the parameter which is used to show the structural foundations for each specimen of sand so that the method and speed of compression would be measured.
Relative density is the mass per unit volume of each specimen of sand which is measured and it has to do with the relative ratio of the density of the sand.
Unit weight is the the exact weight per cubic foot of the sand which is measured.
Please note that your question is incomplete so I gave you a general overview to help you better understand the concept
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Explain mathematically whether the divergence of the curl of a vector field is
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Answer:
yes
Explanation:
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Answers
The water surface elevation must be 110.2631 meters for a flow of 0.03m³ to occur in the cast pipe
For cast iron the chart has 0.0012 from Moody's chart
0.016 for cast iron
[tex]hf = \frac{flQ^{2}}{12.1d5}[/tex]
[tex]h1 = h2+\frac{Q^{2} }{12.1} [\frac{0.0012*100}{(12/1000)^5} +\frac{0.0016*150}{(15/100)^5} ][/tex]
[tex]h1 = 110m+\frac{0.03^2}{12.1} [\f\frac{0.12}{0.00032} +\frac{0.24}{0.000759} ]\\\\h1 = 110+0.0000744[375+3162.06]\\\\= 110 + 0.2631m\\\\= 110.2631m[/tex]
Therefore the water surface elevation must be 110.2631 meters for a flow of 0.03m³ to occur in the cast pipe
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I want to know if anyone else know how to tell apart real leather from fake leather?!?!
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Answer:
yes i do.
Explanation:
it's pretty easy in my opinion but i dont know about you why do you ask?
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a
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C.
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d.
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Answers
Answer:
d.
50.5" is the answer for this question
Consider a 25-mm-diameter and 15-m-long smooth tube that is maintained at a constant surface temperature. Fluids enter the tube at 50°C with a mass flow rate of 0.01 kg/s. Determine the tube surface temperatures necessary to heat water, engine oil, and liquid mercury to the desired outlet temperature of 150°C
Answers
Answer: ⚠this is not my answer it from hamzaahmeds⚠
Water: h = 35.53 W/m².k
Engine oil: h = 18.84 W/m².k
Mercury: h = 1.19 W/m².k
Explanation:
Assuming the steady state, one-dimensional heat flow, it is clear that the added to the fluid by tube heat will be equal to the heat transfer through convection outside the tube.
Therefore,
mCΔT = hAΔT
mC = hA
h = mC/A
where,
h = convection coefficient
m = mass flow rate = 0.01 kg/s
C = specific heat capacity of fluid
A = surface area of tube = 2πrL = 2π(0.0125 m)(15 m) = 1.178 m²
FOR WATER:
C = 4186 J/Kg.k
Therefore,
h = (0.01 kg/s)(4186 J/Kg.k)/(1.178 m²)
h = 35.53 W/m².k
FOR ENGINE OIL:
C = 2220 J/Kg.k
Therefore,
h = (0.01 kg/s)(2220 J/Kg.k)/(1.178 m²)
h = 18.84 W/m².k
FOR LIQUID MERCURY:
C = 140 J/Kg.k
Therefore,
h = (0.01 kg/s)(140 J/Kg.k)/(1.178 m²)
h = 1.19 W/m².k
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Answer:
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Explanation:
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true/False
1.
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Answer:true
Explanation:
It is true that around 50 percent of all sales of farm commodities are from dairy products.
What are dairy products?Dairy products or milk products, also known as lacticinia, are milk-based foods.
Cows, water buffalo, nanny goats, and ewes are the most common dairy animals. Dairy products include common grocery store foods such as yogurt, cheese, and butter in the Western world.
Milk, yogurt, cheese, lactose-free milk, and fortified soy milk and yogurt are all part of the Dairy Group. The Dairy Group excludes milk-based foods with low calcium and a high fat content. Cream cheese, sour cream, cream, and butter are all examples of this.
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Thus, the given statement is true.
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i dob't know
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So, I act like I believe them.
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affects the flow of electricity
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What is one of the reasons to consider implementing a blockchain solution?
Options:
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2. ability to work on different versions of the same data
3.ability to verify that data has not been tampered with since its creation
4.inability to trace and access data of other users
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Answer:
not sure
Explanation:
not sure
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Answer:
Tech B
Explanation:
I took the test and got it right
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Answers
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In Example we assumed that we were going to use all of the solvent available. There are other alternatives. Determine if the following alternatives are capable of producing outlet water of the desired acetic acid concentration.
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Answer:
NO!
Explanation:
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